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Suppose that a sequence of independent tosses is made with a coin for which the probability of obtaining a head-on each given toss is \(\frac{{\bf{1}}}{{{\bf{30}}}}.\)

a. What is the expected number of tails that will be obtained before five heads have been obtained?

b. What is the variance of the number of tails that will be obtained before five heads have been obtained?

Short Answer

Expert verified

a.145

b. 4350

Step by step solution

01

Given information

Let X show the number of tails until the first fiveheadsare obtained.

Here, X is a random variable that follows negative binomial distribution that is

\(X \sim NB\left( {r = 5,p = \frac{1}{{30}}} \right)\).

02

Defining the pdf

This negative binomial distribution is based on failures.

Thus, the pdf is,

\(f\left( x \right) = {}^{x + r - 1}{C_{r - 1}}{p^r}{\left( {1 - p} \right)^{x - r}},r = 0,1,2 \ldots \)

03

(a) Expectation finding

Since X follows a negative binomial distribution, by its properties, the expectation is:

\(\begin{array}{c}E\left( X \right) = \frac{{r\left( {1 - p} \right)}}{p}\\ = \frac{{5\left( {1 - \frac{1}{{30}}} \right)}}{{\frac{1}{{30}}}}\\ = 145\end{array}\)

As a result, the predicted number of tails gained prior to obtaining five heads is 145.

04

(b) Variance finding

Since X follows a negative binomial, by its properties, the variance is:

\(\begin{array}{c}V\left( X \right) = \frac{{r\left( {1 - p} \right)}}{{{p^2}}}\\ = \frac{{5\left( {1 - \frac{1}{{30}}} \right)}}{{\frac{1}{{{{30}^2}}}}}\\ = 4350\end{array}\)

As a result, the variation of the number of tails acquired before five tails are achieved is 4350.

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Most popular questions from this chapter

Suppose that the measured voltage in a certain electric circuit has the normal distribution with mean 120 and standard deviation 2. If three independent measurements of the voltage are made, what is the probability that all three measurements will lie between 116 and 118?

Let \({{\bf{X}}_{{\bf{1,}}}}{\bf{ }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)be i.i.d. random variables having thenormal distribution with mean \({\bf{\mu }}\) and variance\({{\bf{\sigma }}^{\bf{2}}}\). Define\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {{{\bf{X}}_{\bf{i}}}} \) , the sample mean. In this problem, weshall find the conditional distribution of each \({{\bf{X}}_{\bf{i}}}\)given\(\overline {{{\bf{X}}_{\bf{n}}}} \).

a.Show that \({{\bf{X}}_{\bf{i}}}\)and\(\overline {{{\bf{X}}_{\bf{n}}}} \) have the bivariate normal distributionwith both means \({\bf{\mu }}\), variances\({{\bf{\sigma }}^{\bf{2}}}{\rm{ }}{\bf{and}}\,\,\frac{{{{\bf{\sigma }}^{\bf{2}}}}}{{\bf{n}}}\),and correlation\(\frac{{\bf{1}}}{{\sqrt {\bf{n}} }}\).

Hint: Let\({\bf{Y = }}\sum\limits_{{\bf{j}} \ne {\bf{i}}} {{{\bf{X}}_{\bf{j}}}} \).

Nowshow that Y and \({{\bf{X}}_{\bf{i}}}\) are independent normals and \({{\bf{X}}_{\bf{n}}}\)and \({{\bf{X}}_{\bf{i}}}\) are linear combinations of Y and \({{\bf{X}}_{\bf{i}}}\) .

b.Show that the conditional distribution of \({{\bf{X}}_{\bf{i}}}\) given\(\overline {{{\bf{X}}_{\bf{n}}}} {\bf{ = }}\overline {{{\bf{x}}_{\bf{n}}}} \)\(\) is normal with mean \(\overline {{{\bf{x}}_{\bf{n}}}} \) and variance \({{\bf{\sigma }}^{\bf{2}}}\left( {{\bf{1 - }}\frac{{\bf{1}}}{{\bf{n}}}} \right)\).

Let X and P be random variables. Suppose that the conditional distribution of X given P = p is the binomial distribution with parameters n and p. Suppose that the distribution of P is the beta distribution with parameters

\({\bf{\alpha }} = {\rm{ }}{\bf{1}}{\rm{ }}{\bf{and}}\,\,{\bf{\beta }} = {\rm{ }}{\bf{1}}\). Find the marginal distribution of X.

Suppose that two random variables \({{\bf{X}}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{X}}_{\bf{2}}}\)have a bivariate normal distribution, and two other random variables \({{\bf{Y}}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{Y}}_{\bf{2}}}\)are defined as follows:

\(\begin{array}{*{20}{l}}{{{\bf{Y}}_{\bf{1}}}{\rm{ }} = {\rm{ }}{{\bf{a}}_{{\bf{11}}}}{{\bf{X}}_{\bf{1}}}{\rm{ }} + {\rm{ }}{{\bf{a}}_{{\bf{12}}}}{{\bf{X}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{{\bf{b}}_{\bf{1}}},}\\{{{\bf{Y}}_{\bf{2}}}{\rm{ }} = {\rm{ }}{{\bf{a}}_{{\bf{21}}}}{{\bf{X}}_{\bf{1}}}{\rm{ }} + {\rm{ }}{{\bf{a}}_{{\bf{22}}}}{{\bf{X}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{{\bf{b}}_{\bf{2}}},}\end{array}\)

where

\(\left| {\begin{array}{*{20}{l}}{{{\bf{a}}_{{\bf{11}}}}{\rm{ }}{{\bf{a}}_{{\bf{12}}}}}\\{{{\bf{a}}_{{\bf{21}}}}{\rm{ }}{{\bf{a}}_{{\bf{22}}}}}\end{array}} \right|\)

Show that \({{\bf{Y}}_{\bf{1}}}{\rm{ }}{\bf{and}}{\rm{ }}{{\bf{Y}}_{\bf{2}}}\) also have a bivariate normal distribution.

Suppose that a pair of balanced dice are rolled 120 times, and let X denote the number of rolls on which the sum of the two numbers is 12. Use the Poisson approximation to approximate \({\bf{Pr}}\left( {{\bf{X = 3}}} \right){\bf{.}}\)

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